JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 19)
The largest value of the non-negative integer a for which $$\mathop {\lim }\limits_{x \to 1} {\left\{ {{{ - ax + \sin (x - 1) + a} \over {x + \sin (x - 1) - 1}}} \right\}^{{{1 - x} \over {1 - \sqrt x }}}} = {1 \over 4}$$ is
Answer
0
Explanation
Given, $$\mathop {\lim }\limits_{x \to 1} {\left\{ {{{\sin (x - 1) + a(1 - x)} \over {(x - 1) + \sin (x - 1)}}} \right\}^{{{(1 - \sqrt x )(1 - \sqrt x )} \over {1 - \sqrt x }}}} = {1 \over 4}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {\left\{ {{{{{\sin (x - 1)} \over {(x - 1)}} - a} \over {1 + {{\sin (x - 1)} \over {(x - 1)}}}}} \right\}^{1 + \sqrt x }} = {1 \over 4}$$
$$ \Rightarrow {\left( {{{1 - a} \over 2}} \right)^2} = {1 \over 4}$$
$$ \Rightarrow {(a - 1)^2} = 1$$
$$\Rightarrow$$ a = 2 or 0
But for a = 2, base of above limit approaches $$-$$1/2 and exponent approaches to 2 and since base cannot be negative, hence limit does not exist.
$$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {\left\{ {{{{{\sin (x - 1)} \over {(x - 1)}} - a} \over {1 + {{\sin (x - 1)} \over {(x - 1)}}}}} \right\}^{1 + \sqrt x }} = {1 \over 4}$$
$$ \Rightarrow {\left( {{{1 - a} \over 2}} \right)^2} = {1 \over 4}$$
$$ \Rightarrow {(a - 1)^2} = 1$$
$$\Rightarrow$$ a = 2 or 0
But for a = 2, base of above limit approaches $$-$$1/2 and exponent approaches to 2 and since base cannot be negative, hence limit does not exist.
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